(a) You have a stock solution of \(14.8 \mathrm{M} \mathrm{NH}_{3}\). How many milliliters of this solution should you dilute to make \(1000.0 \mathrm{~mL}\) of \(0.250 \mathrm{M} \mathrm{NH}_{3} ?\) (b) If you take a \(10.0-\mathrm{mL}\) portion of the stock solution and dilute it to a total volume of \(0.500 \mathrm{~L},\) what will be the concentration of the final solution?

Using the dilution formula (\(C_1V_1 = C_2V_2\)), we can plug in the known values for initial concentration (\(C_1 = 14.8\mathrm{M}\)), final concentration (\(C_2 = 0.250\mathrm{M}\)), and final volume (\(V_2 = 1000.0\mathrm{mL}\)). Then, we solve for initial volume (\(V_1\)): \(14.8\mathrm{M} \times V_1 = 0.250\mathrm{M} \times 1000.0\mathrm{mL}\) To find the value of \(V_1\), we divide both sides of the equation by \(14.8\mathrm{M}\): \(V_1 = \frac{0.250\mathrm{M} \times 1000.0\mathrm{mL}}{14.8\mathrm{M}}\) Now, we can calculate the value of \(V_1\): \(V_1 = \frac{0.250 \times 1000.0}{14.8} = 16.8919\mathrm{mL}\) Therefore, we need to dilute \(16.8919\mathrm{mL}\) of the stock solution to make \(1000.0\mathrm{mL}\) of \(0.250\mathrm{M} \mathrm{NH}_{3}\).

For part (b), we have the initial concentration (\(C_1 = 14.8\mathrm{M}\)), initial volume (\(V_1 = 10.0\mathrm{mL}\)), and the final volume (\(V_2 = 0.500\mathrm{L} = 500.0\mathrm{mL}\)). We can plug these values into the dilution formula to find the final concentration (\(C_2\)): \(14.8\mathrm{M} \times 10.0\mathrm{mL} = C_2 \times 500.0\mathrm{mL}\) To solve for \(C_2\), we divide both sides of the equation by \(500.0\mathrm{mL}\): \(C_2 = \frac{14.8\mathrm{M} \times 10.0\mathrm{mL}}{500.0\mathrm{mL}}\) Now, we can calculate the value of \(C_2\): \(C_2 = \frac{14.8 \times 10.0}{500.0} = 0.296\mathrm{M}\). Therefore, the concentration of the final solution after dilution is \(0.296\mathrm{M}\).